. If the condition m(t) ≤1 is not satisfied and thepercentage of negative modulation is over 100%,theenvelope detector can not be used. Ex. Power of an AM signal (description of the question)AM broadcast transmitter:a 5000-W transmitter is connectedto a 50ohms load;then the constant A. is given by1/2A.2/50=5000.So the peak voltage across the load will beA =707V during the times of no modulation.If thetransmitter is then 100%modulated by a 100-Hz test tone,the total (carrier plus sideband) average power will be :1.5[1/2(A/50)]=7500WThere we have <m?(t)>=1/2 for a sinusoidal modulationwaveshape of unity (100%) amplitude.The modulation efficiency would be 33% since <m?(t)>=1/2
• If the condition │m(t)│≤1 is not satisfied and the percentage of negative modulation is over 100%,the envelope detector can not be used. • Ex. Power of an AM signal (description of the question) AM broadcast transmitter:a 5000-W transmitter is connected to a 50ohms load;then the constant Ac is given by 1/2Ac 2 /50=5000.So the peak voltage across the load will be Ac =707V during the times of no modulation.If the transmitter is then 100%modulated by a 100-Hz test tone, the total (carrier plus sideband) average power will be : 1.5[1/2(Ac 2 /50)]=7500W There we have <m2 (t)>=1/2 for a sinusoidal modulation waveshape of unity (100%) amplitude. The modulation efficiency would be 33% since <m2 (t)>=1/2
Linear modulation----Double-Sideband suppressedcarrier modulation (DSB)Mapping function:gDsB[-]=A。 :Modulated waveform:s(t)=Re(Am(t)l ejoct)=Acm(t)cos0ctSpectrum:S(f)=1/2A.[M(f+f.)+M(f-f.)]Normalized average power of s(t):<s?(t)>=1/2A2<m?(t)>Diagram ofDSB system:BPFLPFchannelm(t)s(t)m'(t)s'(t)+n(t)Accosoctcosoct
• Linear modulation-Double-Sideband suppressed carrier modulation (DSB) • Mapping function:gDSB[ . ]=Ac . • Modulated waveform: s(t)=Re{Acm(t)] ejωc t}=Acm(t)cosωc t • Spectrum:S(f)=1/2Ac[M(f+fc)+M(f-fc)] • Normalized average power of s(t): <s2 (t)>=1/2Ac 2<m2 (t)> • Diagram of DSB system: A s’(t)+n(t) ccosωct channel m(t) s(t) BPF cosωct LPF m’(t)
.Linear modulation----Single-Sideband modulation(SSB)Definition:An upper sideband (USSB) signal has a zero-valued spectrum for f<f, where f。 is the carrierfrequency.A lower sideband (LSSB) signal has a zero-valuedspectrum for I f>f., where f. is the carrier frequency.: Mapping function:g ssb[m(t)] = A.[m(t) ± m(t)]Modulated waveform:s(t)-Re{Acg(t)] ejoct)=Ac[m(t)cosoct m (t) sinoct]where the upper (-) sign is used for USSB and the lower (+)is for LSSB. m^(t) denotes the Hilbert transform of m(t)m^(t)=m(t)*h(t)
• Linear modulation-Single-Sideband modulation (SSB) • Definition:An upper sideband (USSB) signal has a zerovalued spectrum for │f│<fc , where fc is the carrier frequency. A lower sideband (LSSB) signal has a zero-valued spectrum for │f│>fc , where fc is the carrier frequency. • Mapping function: • Modulated waveform: s(t)=Re{Acg(t)] ejωc t}=Ac[m(t)cosωc t m^(t) sinωc t] where the upper (-) sign is used for USSB and the lower (+) is for LSSB. m^(t) denotes the Hilbert transform of m(t). m^(t)=m(t)*h(t) [ ( )] [ ( ) ( )] g m t = A m t m t SSB c
h(t)=1 /元tandH(f)=-f for f>0 and H(f)=f for f<0M(f)t/ G(f) I2Ac1fBBfUSSBI s() IAcff.f.+B-fc-B-fc
h(t)=1/πt and H(f)=-f for f>0 and H(f)=f for f<0 M(f) B f 1 │G(f)│ B 2Ac f USSB │S(f)│ f fc Ac -fc-B -fc fc+B